One third of a number increased by twice the number

Compare 8 ⋅ 107 to 2 ⋅ 104. Plz Answer 25 Points!!!

Shalnov [3]

The answer is the first row.

Explanation:

8 * 10^7 = 80,000,000 -- 2 * 10^4 = 20,000

80,000,000 divided by 20,000 = 4,000, so it is 4,000 times larger

Therefore, the first answer is correct.

Hope I helped and good luck!

3 0

3 years ago

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Grace is three times as old as Hans, but in 5 years she will be twice as old as Hans is then. How old are they now? Set up an th

frosja888 [35]

Answer:

3x - y = 0; 2x - y = -5

Step-by-step explanation:

Let x be the present age of Hans and y be the present age of Grace,

Since, in present Grace is three times as old as Hans,

⇒ y = 3x

⇒ 3x - y = 0

Now, after 5 years,

The age of Hans = x + 5,

And, the age of Grace = y + 5

Also, in 5 years Grace will be twice as old as Hans is then,

⇒ y + 5 = 2 ( x + 5 )

⇒ y + 5 = 2x + 10

⇒ 2x - y = -5

Hence, the required system of linear equations is,

3x - y = 0; 2x - y = -5

5 0

3 years ago

Round your answer to the nearest hundredth.<br><br> 13.48x − 200 < 256.12

wolverine [178]

Answer:

Step-by-step explanation:

13.48x − 200 < 256.12

Using the addition property , add 200 to both sides

That's

13.48x + 200 - 200 < 256.12 + 200

13.48x < 456.12

<u>Divide both sides by 13.48</u>

$\frac{13.48x}{13.48} < \frac{456.12}{13.48} \\ x < \frac{456.12}{13.48} \\ \: \: \: x < 33.83679$

We have the final answer as

x < 33.84 to the nearest hundredth

Hope this helps you

4 0

3 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago

Explain how find is 7 x 20 is similar to find 7 X 2000 product

Harman [31]

You just multiply 7 x 20 and you will get the answer of 140 then just add two zeroes to get 14000.

4 0

3 years ago

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